Let the straight line touch a circle with center , , and radius at a point . Let be the point on the circle such that the line segment is a diameter of the circle. Let . [2024]

Question

Let the straight line $y = 2 x$ touch a circle with center $(0, α)$ , $α > 0$ , and radius $r$ at a point $A_{1}$ . Let $B_{1}$ be the point on the circle such that the line segment $A_{1} B_{1}$ is a diameter of the circle. Let $α + r = 5 + \sqrt{5}$ .

Match each entry in List-I to the correct entry in List-II.

	List-I		List-II
(P)	$α$ equals	(1)	$(- 2, 4)$
(Q)	$r$ equals	(2)	$\sqrt{5}$
(R)	$A_{1}$ equals	(3)	$(- 2, 6)$
(S)	$B_{1}$ equals	(4)	$5$
		(5)	$(2, 4)$

The correct option is [2024]

Smart Achievers · Accepted Answer

(3)

$Consider centre as P (0, α), α > 0$

$Distance of A_{1} P =$ $| \frac{2 (0) - α}{\sqrt{5}} |$ $= r$

$\Rightarrow$ $| - α |$ $= \sqrt{5} r \Rightarrow α = \sqrt{5} r$ $∴$ $α + r = 5 + \sqrt{5}$

$\Rightarrow \sqrt{5} r + r = \sqrt{5} (\sqrt{5} + 1) \Rightarrow r = \sqrt{5}, α = 5$

$∴$ $P (0, 5)$

$Foot of perpendicular from P to line 2 x - y = 0$

$\frac{x - 0}{2} = \frac{y - 5}{- 1} = \frac{- (2 (0) - 5)}{5} = 1$

$\Rightarrow x = 2, y = 4 A_{1} (2, 4)$

$Let B (x_{1}, y_{1}),$ $∴$ $\frac{x_{1} + 2}{2} = 0, \frac{y_{1} + 4}{2} = 5$

$\Rightarrow x_{1} = - 2, y_{1} = 6 B (- 2, 6)$

Solution

1 (P) → (4), (Q) → (2), (R) → (1), (S) → (3)

2 (P) → (2), (Q) → (4), (R) → (1), (S) → (3)

3 (P) → (4), (Q) → (2), (R) → (5), (S) → (3)

4 (P) → (2), (Q) → (4), (R) → (3), (S) → (5)

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